"An urn contains one white chip and a second chip that is equally likely to be white or black. A chip is drawn at random and returned to the urn. Then a second chip is drawn. What is the probability that a white appears on the second draw given that a white appeared on the first draw"?
I keep getting answer 3/4 but it is incorrect.
I tried calculating P(A|B) directly; also, I looked at the case where both chips were white then one was black and I averaged those. I always got 3/4.
Personally, I don't even see how the first drawing has any affect on the latter.
Best Answer
Conditional probabilities are not a matter of cause and effect; see also this answer I just gave. The question is not whether the first draw has any effect on the second one, but whether your knowledge of the result of the first draw makes a difference for the probabilities you assign to the outcomes of the second one.
If $A$ is the event of drawing white on the second draw and $B$ is the event of drawing white on the first draw, then, since there is an equal chance of $1/2$ each of the second chip being white or black,
$$P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{\frac12\cdot1+\frac12\cdot\frac14}{\frac12\cdot1+\frac12\cdot\frac12}=\frac{1+\frac14}{1+\frac12}=\frac56\;.$$